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  • [luoguP1273] 有线电视网(DP)

    传送门

    f[i][j]表示节点i选j个用户的最大收益

    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #define N 3001
    #define max(x, y) ((x) > (y) ? (x) : (y))
    
    int n, m, cnt;
    int head[N], to[N], next[N], val[N], f[N][N], size[N];
    //f[i][j]表示第i个节点选择j个用户的最优解 
    
    inline int read()
    {
    	int x = 0, f = 1;
    	char ch = getchar();
    	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
    	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
    	return x * f;
    }
    
    inline void add(int x, int y, int z)
    {
    	to[cnt] = y;
    	val[cnt] = z;
    	next[cnt] = head[x];
    	head[x] = cnt++;
    }
    
    inline void dfs(int u)
    {
    	int i, j, k, v;
    	size[u] = 1;
    	for(i = head[u]; i ^ -1; i = next[i])
    	{
    		v = to[i];
    		dfs(v);
    		size[u] += size[v];
    		for(j = size[u]; j >= 1; j--)
    			for(k = 1; k <= j; k++)
    				f[u][j] = max(f[u][j], f[u][j - k] + f[v][k] - val[i]);
    	}
    }
    
    int main()
    {
    	int i, j, k, x, y;
    	n = read();
    	m = read();
    	memset(head, -1, sizeof(head));
    	for(i = 1; i <= n - m; i++)
    	{
    		k = read();
    		for(j = 1; j <= k; j++)
    		{
    			x = read();
    			y = read();
    			add(i, x, y);
    		} 
    	}
    	for(i = 1; i <= n; i++)
    		for(j = 1; j <= n; j++)
    			f[i][j] = -10000;
    	for(i = n - m + 1; i <= n; i++) f[i][1] = read();
    	dfs(1);
    	for(i = m; i >= 0; i--)
    		if(f[1][i] >= 0)
    		{
    			printf("%d
    ", i);
    			return 0;
    		}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zhenghaotian/p/7344173.html
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